Triangle Median Theorem proof

Question

Does anyone know what theorem this is? Or how to prove it?

Let $ABC$ be a triangle with sides $a, b, c$. Let $D$ be the midpoint of side $BC$ and $w = AD$ the length of the median through $A$. Prove that: $$w^2={b^2\over 2}+{c^2\over 2}-{a^2\over 4}.$$

Answer 1

Reflect $A$ across $D$, we get new point $E$ such that $ABEC$ is parallelogram.

By the Parallelogram Identity (which can be easily proved, say by the The Law of Cosine) we have $$2AB^2+2AC^2 = AE^2+BC^2$$

$$2c^2+2b^2 = a^2+(2w)^2$$

Now you can finish...

Answer 2

We can also prove it using coordinate geometry.

Let the coordinates of point B be (0, 0) (Without Loss of Generality), coordinates of C be (a, 0) (Without Loss of Generality) and coordinates of A be (n, m).

$$AB = c$$ $$BC = a$$ $$AC = b$$ $$AD = w$$

$$AB² = c² = [(n-0)² + (m-0)²]$$ $$c² = n² + m²$$ $$BC² = a² = [(a-0)² + (0-0)²]$$ $$a² = a²$$ $$AC² = b² = [(n-a)² + (m-0)²]$$ $$b² = n² + a² - 2an + m²$$ $$AD² = w² = [(n-(a/2))² + (m-0)²]$$ $$w² = (1/4)[4n² + a² - 4an + 4m²]$$

$$(b²/2) + (c²/2) - (a²/4) = (1/4)[2b² + 2c² - a²]$$ $$ = (1/4)[ 2n² + 2a² - 4an + 2m² + 2n² + 2m² - a²]$$ $$ = (1/4)[4n² + a² - 4an + 4m²]$$ $$ = w²$$